Question

(6+x)^2=(x-2)(3-x) , X(8-3x)=(5x-1)^2 приведите уравнение к виду ax^2+bx+c=0ПОМОГИТЕ ПОЖАЛУЙСТА!!!!!!

Answer 1

(6+x)²=(x-2)(3-x)
36+x²=3x-6-x²+2x
x²+x²-5x+36+6=0
2x²-5x+42=0

x(8-3x)=(5x-1)²
8x-3x²=25x²-10x+1
25x²+3x²-10x-8x+1=0
28x²-18x+1=0

Answer 2

Equation 1: (6+x)^2 = (x-2)(3-x)

To solve the equation (6+x)^2 = (x-2)(3-x), we can expand both sides and simplify.

Expanding the left side of the equation: (6+x)^2 = (6+x)(6+x) = 36 + 12x + x^2

Expanding the right side of the equation: (x-2)(3-x) = 3x - x^2 - 6 + 2x = -x^2 + 5x - 6

Now we have the equation: 36 + 12x + x^2 = -x^2 + 5x - 6

To solve this quadratic equation, we can move all the terms to one side to get: 2x^2 + 7x - 42 = 0

The equation is now in the form ax^2 + bx + c = 0, where a = 2, b = 7, and c = -42.

Equation 2: X(8-3x) = (5x-1)^2

To solve the equation X(8-3x) = (5x-1)^2, we can expand both sides and simplify.

Expanding the left side of the equation: X(8-3x) = 8X - 3X^2

Expanding the right side of the equation: (5x-1)^2 = (5x-1)(5x-1) = 25x^2 - 10x + 1

Now we have the equation: 8X - 3X^2 = 25x^2 - 10x + 1

To solve this quadratic equation, we can move all the terms to one side to get: 3X^2 + 17x - 1 = 0

The equation is now in the form ax^2 + bx + c = 0, where a = 3, b = 17, and c = -1.

Please note that the solutions to these equations can be found by using various methods such as factoring, completing the square, or using the quadratic formula. If you need further assistance in solving these equations, please let me know.